Re: Firefox XSLT position()

David wrote:


> So either I put my XML on one line something like :
> <AAA><BBB>a</BBB><BBB>b</BBB><BBB>c</BBB></AAA>

If you are only interested in child elements then as you have already 
found out you can use
   <xsl:apply-templates select="*" />
instead of
   <xsl:apply-templates />
that way both XSLT processors process the same nodeset and position() 
will yield the same results.

-- 

	Martin Honnen
	http://JavaScript.FAQTs.com/